complex-analysiscontour-integrationsolution-verification
Consider the path $ γ:[2π,5π]→ \mathbb{C} $ given by $ γ(t)=te^{it}. $
what is $$ \int_\gamma \frac{1}{z} dz $$
Using the definition of integral of a complex function over a path?
I got as an answer: $ log \frac{5}{2} +3\pi i $ but aparently that's wrong. I used the integral $ \int_{2\pi}^{5\pi} $ but in the solution they used $\int_{\pi}^{4\pi} $. Why is that?
I thought about this once and came up with an "answer" of sorts.
Notation: Given $z=a+ib$ and $w = c+id$ in $\mathbb{C}$, I will write $z \cdot w$ for the Euclidean dot product $ac + bd$. Ordinary multiplication of complex numbers will be denoted by juxtaposition.
The basic observation is, given $z,w \in \mathbb{C}$, the quantity $\overline z w$ is related to the dot product as follows: \begin{align*} \mathrm{Re}(\overline z w) = z \cdot w && \mathrm{Im}(\overline z w) = (iz) \cdot w. \end{align*} Note that $iz$, geometrically speaking, is $z$ rotated counterclockwise by 90 degrees.
Now, rather than interpret $\int_\gamma f$, let us interpret $I = \int_\gamma \overline f$. As you noted, $$ I= \int_0^1 \overline{f(\gamma(t))} \gamma'(t) \ dt,$$ and so \begin{align*} \mathrm{Re}(I) = \int_0^1 f(\gamma(t)) \cdot \gamma'(t) \ dt && \mathrm{Im}(I) = \int_0^1 (if(\gamma(t))) \cdot \gamma'(t) \ dt \end{align*} or, in the notation of line integrals of vector fields, \begin{align*} \mathrm{Re}(I) = \int_\gamma f(z) \cdot dz && \mathrm{Im}(I) = \int_\gamma (if(z)) \cdot dz. \end{align*} So, for instance, thinking of $f$ as a force field,
Note $if$ is just the force field $f$ rotated counterclockwise by 90 degrees.
Hope this helps...
Added: I thought I would add a simple example.
Put $f(z) = \overline{1/z}$. As you are no doubt aware, if the contour $\gamma$ is closed and avoids the origin, then $\int_\gamma \overline f = \int_\gamma \frac{1}{z}$ is equal to $2 \pi i$ times the winding number of $\gamma$.
In this example, the vector field $f(z) = \overline{1/z}$ is the gradient of the scalar potential $V(x,y) = \frac{1}{2} \log (x^2+y^2)$, so no work is done going around a contour in this field. The field $f$ looks like this:
On the other hand, the 90 degree rotated field $if$ looks like this:
Added: I just learned the proper name for this interpretation. The complex conjugate $\overline f$, viewed as a vector field, is called the "Polya vector field of $f$". As I mentioned, the real part of $\int_\gamma f$ is the work done by the Polya vector vield on a particle as it travels along the curve $\gamma$. An alternative way to think of the imaginary part of $\int_\gamma f$ is as the flux of the Polya vector field through the oriented curve $\gamma$. These ideas are explained in Chapter 11 of Tristan Needham's book "Visual Complex Analysis".
In this case no parametrization is required, because the function $z\mapsto\frac1{(1-z)^2}$ has a primitive: $z\mapsto\frac1{1-z}$. So, for any differentibale path $\gamma\colon[a,b]\longrightarrow\Bbb C\setminus\{1\}$, you have$$\int_\gamma\frac1{(1-z)^2}\,\mathrm dz=\frac1{1-u(b)}-\frac1{1-u(a)}.$$In general, if $f$ has a primitive $F$, then\begin{align}\int_\gamma f&=\int_a^bf\bigl(\gamma(t)\bigr)\gamma'(t)\,\mathrm dt\\&=\int_a^bF'\bigl(\gamma(t)\bigr)\gamma'(t)\,\mathrm dt\\&=\int_a^b(f\circ\gamma)'(t)\,\mathrm dt\\&=F\bigl(\gamma(b)\bigr)-F\bigl(\gamma(a)\bigr).\end{align}
And, yes, you can differentiate your $\gamma$ the way you did it.
Best Answer
Since\begin{align}\int_\gamma\frac1z\,\mathrm dz&=\int_{2\pi}^{5\pi}i+\frac1t\,\mathrm dt\\&=\log\left(\frac52\right)+3\pi i,\end{align}your answer is correct. Since $\gamma$ is not periodic, it makes no sense to use $[\pi,4\pi]$ instead of $[2\pi,5\pi]$.